# Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$

Let $$a,b,c\in\mathbb{Z}$$, $$1, $$c$$ is a prime number and $$f(x)=ax^2+bx+c$$. If $$f(f(1))=f(f(2))=f(f(3))$$, find $$f'(f(1))+f(f'(2))+f'(f(3))$$

My attempt: \begin{align*} f'(x)&=2ax+b\\ (f(f(x)))'&=f'(f(x))f'(x)\\ f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\ \end{align*}

• Clarification : in the " find ... " part, is it $f'(f(2))$ or $f(f'(2))$? Jan 7, 2021 at 7:09
• @TeresaLisbon It's $f(f'(2))$.
– Ken
Jan 7, 2021 at 7:12
• Thank you for the clarification. Jan 7, 2021 at 7:12
• In case that gives you some insight, Mathcad reduces your problem to $b=-4a$ and $c=2+\frac{7a}{2}$. Enumerating $a$ from $2$ to $9$ yields you the only prime $c$ that gives an integer $b$, which means $(a,b,c)=(6, -24, 23)$, and finally $f'(f(1))+f(f'(2))+f'(f(3))=95$. Quick brute-force search through $c \in (first \; 1000 \; primes)$ and $-10000<b<10000$ returns no other valid solutions. Jan 7, 2021 at 9:29
• @Thehx How is $c=2+7a/2$ obtained? Jan 7, 2021 at 9:58

Due to the symmetry of vertical parabola, for distinct $$x_1, x_2,$$ $$f(x_1)=f(x_2) \Rightarrow x_1+x_2=-b/a$$ and $$f'(x_1)+f'(x_2)=0$$

For a quadratic, w cannot have $$f(x_1)=f(x_2)=f(x_3)$$ for distinct three $$x_i$$ since this would imply, $$f(x)−f(x_1)$$ has three distinct roots. There arise three conditions of arguments being pairwise equal.

$$f(f(1))=f(f(2))=f(f(3)) \Rightarrow$$ following cases :

• $$f(1)=f(2) \Rightarrow 1+2=-b/a$$.

Now $$f(1) \neq f(3)$$. But $$f(f(1))=f(f(3))$$. $$\Rightarrow f(1)+f(3)=-b/a=3$$

Here $$f(1)+f(2) \neq -b/a$$ since $$f(1),f(2)$$ are identical. Similarly,

• $$f(2)=f(3) \Rightarrow 2+3=-b/af(1) \neq f(3) \Rightarrow f(1)+f(3)=-b/a=5$$
• $$f(1)=f(3) \Rightarrow 1+3=-b/af(1) \neq f(2) \Rightarrow f(1)+f(2)=-b/a=4$$

But $$f(1)+f(3)=10a+4b+2c$$ is even. So only the third case holds: $$f(1)=f(3)$$ and $$f'(1)+f'(3)=0$$.

Thus $$b=-4a$$. $$f(x)=ax^2-4ax+c \Rightarrow f'(x)=2a(x-2)$$

Also $$f'(f(x)) = (ay^2-4ay+c)'=2a(y-2)y'=2af'(x)(f(x)-2)$$

Evaluating, $$f'(f(1))+f(f'(2)+f'(f(3))=c$$

where $$a,c$$ have been computed as below.

Remark :

Inspired from the other answer, even the numerical value of $$c$$ can be calculated. Subbing $$f(x)=ax^2-4ax+c$$ into $$f(f(1))=f(f(2))$$, one obtains $$c=\dfrac{7a+4}{2}$$

Quickly checking for $$a\in \{2,4,6,8\}$$, $$c=23$$ a prime, only for $$a=6$$.

• I believe your statement about $f'(f(1))+f(f'(2))+f'(f(3))=c$ is wrong. For $a=6, b=-24, c=23$, said sum equals $95$, not $23$. Other than that, I like how you approached this one, good job! Jan 7, 2021 at 11:11
• @Thehx it's quite possible I made a mistake and am not able to find it - I regularly do. I'll let others also calculate the same and correct me. I have duly upvoted your answer. Jan 7, 2021 at 11:15
• I tried following your steps, but I have questions. (a) from $f(f(1))=f(f(2))$ you jump to $f(1)=f(2)$, which disturbs me since $f$ is not yet proven bijective (well duh, it is not bijective) — and could get us to losing some solutions. Still, if we instead write $f(f(1))=f(f(2)) \Rightarrow f(1)+f(2) = -b/a$, just as you said first, we're good, and will get to the same set of conditions of $b=-Ka$. The last line is wrong or unfinished (did you not copy the entire latex?). The right formula should be $f'(f(1))+f(f'(2))+f'(f(3))=2a^2+7a+2$ Jan 7, 2021 at 11:46
• @Thehx, For a quadratic, one can't have $f(x_1)=f(x_2)=f(x_3)$ for distinct three $x_i$ since this would imply, $f(x)-f(x_1)$ has three roots. So I have taken three conditions of arguments being pairwise equal. Please see the edit. For the last line I believe we're only differing on sign $36+23+36=95$ while $-36+23+36=23$ Jan 7, 2021 at 12:50
• Just please replace $f'(f(1))+f(f'(2)+f'(f(3))=c$ with $f'(f(1))+f(f'(2)+f'(f(3))=2a^2+\frac{7a}{2}+2$. In my earlier comment I forgot to divide $7a$ by two, but please remove that misleading $f'(f(1))+f(f'(2))+f'(f(3))=c$. It does not equal c, period. In terms of $c$ it equals $25c-480$, but not just $c$. Sorry to keep pointing it out. Jan 7, 2021 at 17:45

Okay here's the outline of how you solve this.

first, you write down the system of equations $$0=f(f(2))-f(f(1))=(3a+b)*(b+5a^2+3ab+2ac) \\ 0=f(f(3))-f(f(2))=(5a+b)*(b+13a^2+5ab+2ac)$$ and this leaves you with four possibilities:

(a) $$(3a+b=0)$$ and $$(5a+b=0)$$ which is impossible since $$a \ne 0$$

(b) $$(3a+b=0)$$ and $$(b+13a^2+5ab+2ac=0)$$ in which case we obtain $$b=-3a$$, then our second condition turns into $$2ac-2a^2-3a=0$$, which further yields $$c=a+\frac{3}{2}$$, and we see no acceptable solutions here since both $$a$$ and $$c$$ are asked to be integers.

(c) $$(b+13a^2+5ab+2ac=0)$$ and $$(b+5a^2+3ab+2ac=0)$$ which, if we subtract one of the equations from the other, gives us $$b=-4a$$ and, shortly after, $$p=\frac{7a+4}{2}$$, and we can just check the values of $$a$$ from $$2$$ to $$9$$ to find the only solution at $$a=6, b=-24, c=23$$.

(d) $$(5a+b=0)$$ and $$(b+5a^2+3ab+2ac=0)$$, in which case we get $$b=-5a$$, the second condition simplifies to $$10a-2c+5$$, which gives us $$c=5a+\frac{5}{2}$$, and we also see that $$c$$ will never be integer if $$a$$ is integer, and $$a$$ is always integer.

This leaves us with the only acceptable solution found in (c). As I already mentioned above, putting $$(a,b,c)=(6,-24,23)$$ into $$f'(f(1))+f(f'(2))+f'(f(3))$$ gives you $$95$$.

• Considering one more case $f(3)=f(1) \implies 4a+b=0$ that'll ease your work. Jan 7, 2021 at 10:18
• Yes, $c=23$. But the sum $f'(f(1))+f(f'(2))+f'(f(3))$ equals $95$. Jan 7, 2021 at 10:36